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9702 · 3.2

Non-uniform motion — practice questions

Practice and worked examples for 9702 Non-uniform motion. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A 2.5 kg object accelerates uniformly from rest to 15 m/s in 3.0 seconds due to a constant resultant force. Calculate the magnitude of this resultant force.

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  1. Identify knowns and unknowns: Mass (mm) = 2.5 kg Initial velocity (uu) = 0 m/s (from rest) Final velocity (vv) = 15 m/s Time (tt) = 3.0 s Resultant force (FresultantF_{resultant}) = ?

  2. Calculate the acceleration (aa): Using the kinematic equation: a=vuta = \frac{v - u}{t} a=15 m/s0 m/s3.0 sa = \frac{15 \text{ m/s} - 0 \text{ m/s}}{3.0 \text{ s}} a=5.0 m/s2a = 5.0 \text{ m/s}^2

  3. Apply Newton's Second Law (Fresultant=maF_{resultant} = ma): Fresultant=2.5 kg×5.0 m/s2F_{resultant} = 2.5 \text{ kg} \times 5.0 \text{ m/s}^2 Fresultant=12.5 NF_{resultant} = 12.5 \text{ N}

    The magnitude of the resultant force is 12.5 N.

Worked example 2

An 80.0 kg skydiver jumps from a plane. Assume the acceleration due to gravity, g, is 9.81 m/s². (a) Calculate the skydiver's weight. (b) At an instant during the fall, the air resistance is 600 N. Calculate the resultant downward force and the skydiver's acceleration at this moment. (c) What is the magnitude of the air resistance when the skydiver reaches terminal velocity?

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  1. (a) Calculate the weight: Weight is the force due to gravity. Formula: W=mgW = mg W=(80.0 kg)×(9.81 m/s2)W = (80.0 \text{ kg}) \times (9.81 \text{ m/s}^2) W=784.8 NW = 784.8 \text{ N} The skydiver's weight is 785 N (to 3 s.f.).

  2. (b) Calculate resultant force and acceleration: The forces acting are weight (downwards) and air resistance (upwards). We'll take downwards as the positive direction. Resultant Force (FnetF_{net}) = Weight - Air Resistance Fnet=784.8 N600 NF_{net} = 784.8 \text{ N} - 600 \text{ N} Fnet=184.8 NF_{net} = 184.8 \text{ N} The resultant downward force is 185 N (to 3 s.f.).

    Now, use Newton's Second Law to find acceleration (aa). Formula: Fnet=maF_{net} = ma a=Fnetma = \frac{F_{net}}{m} a=184.8 N80.0 kga = \frac{184.8 \text{ N}}{80.0 \text{ kg}} a=2.31 m/s2a = 2.31 \text{ m/s}^2 The skydiver's acceleration at this instant is 2.31 m/s² downwards.

  3. (c) Air resistance at terminal velocity: By definition, terminal velocity is reached when the resultant force is zero. This means the upward forces balance the downward forces. Air Resistance = Weight Air Resistance = 784.8 N The magnitude of the air resistance at terminal velocity is 785 N (to 3 s.f.).